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5.2: Monotonic Functions


Definition

Suppose $$D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$$ and $$(a, b) \subset D .$$ We say $$f$$ is increasing on $$(a, b)$$ if $$f(x)<f(y)$$ whenever $$a<x<y<b ;$$ we say $$f$$ is decreasing on $$(a, b)$$ if $$f(x)>f(y)$$ whenever $$a<x<y<b ;$$ we say $$f$$ is nondecreasing on $$(a, b)$$ if $$f(x) \leq f(y)$$ whenever $$a<x<y<b ;$$ and we say $$f$$ is nonincreasing on $$(a, b)$$ if $$f(x) \geq f(y)$$ whenever $$a<x<y<b .$$ We will say $$f$$ is monotonic on $$(a, b)$$ if $$f$$ is either nondecreasing or nonincreasing on $$(a, b)$$ and we will say $$f$$ is strictly monotonic on $$(a, b)$$ if $$f$$ is either increasing or decreasing on $$(a, b)$$.

Proposition $$\PageIndex{1}$$

If $$f$$ is monotonic on $$(a, b),$$ then $$f(c+)$$ and $$f(c-)$$ exist for every $$c \in(a, b)$$.

Proof

Suppose $$f$$ is nondecreasing on $$(a, b) .$$ Let $$c \in(a, b)$$ and let

$\lambda=\sup \{f(x): a<x<c\}.$

Note that $$\lambda \leq f(c)<+\infty .$$ Given any $$\epsilon>0,$$ there must exist $$\delta>0$$ such that

$\lambda-\epsilon<f(c-\delta) \leq \lambda .$

Since $$f$$ is nondecreasing, it follows that

$|f(x)-\lambda|<\epsilon$

whenever $$x \in(c-\delta, c) .$$ Thus $$f(c-)=\lambda .$$ A similar argument shows that $$f(c+)=\kappa$$ where

$\kappa=\inf \{f(x): c<x<b\}.$

If $$f$$ is nonincreasing, similar arguments yield

$f(c-)=\inf \{f(x): a<x<c\}$

and

$f(c+)=\sup \{f(x): c<x<b\}.$

Proposition $$\PageIndex{2}$$

If $$f$$ is nondecreasing on $$(a, b)$$ and $$a<x<y<b,$$ then

$f(x+) \leq f(y-).$

Proof

By the previous proposition,

$f(x+)=\inf \{f(t): x<t<b\}$

and

$f(y-)=\sup \{f(t): a<t<y\}.$

Since $$f$$ is nondecreasing,

$\inf \{f(t): x<t<b\}=\inf \{f(t): x<t<y\}$

and

$\sup \{f(t): a<t<y\}=\sup \{f(t): x<t<y\}.$

Thus

$f(x+)=\inf \{f(t): x<t<y\} \leq \sup \{f(t): x<t<y\}=f(y-).$

Q.E.D.

Exercise $$\PageIndex{1}$$

Let $$\varphi: \mathbb{Q} \cap[0,1] \rightarrow \mathbb{Z}^{+}$$ be a one-to-one correspondence. Define $$f:[0,1] \rightarrow \mathbb{R}$$ by

$f(x)=\sum_{q \in \mathbb{Q} \cap[0,1]_{q \leq x}} \frac{1}{2^{\varphi(q)}}.$

a. Show that $$f$$ is increasing on $$(0,1)$$.

b. Show that for any $$x \in \mathbb{Q} \cap(0,1), f(x-)<f(x)$$ and $$f(x+)=f(x)$$.

c. Show that for any irrational $$a, 0<a<1, \lim _{x \rightarrow a} f(x)=f(a)$$.

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